How to show that $e^{tA}=\frac{1}{2\pi i}\int_{\{Re \ \lambda =a\}}e^{\lambda t}(\lambda I-A)^{-1}d\lambda$? Let $X$ be a Banach space and $A:X\to X$ be a bounded operator. We can show that if $|\lambda|>|A|$ then $\lambda I-A$ is invertible and 
$$(\lambda I-A)^{-1}=\sum_{k=0}^{+\infty}\frac{A^k}{\lambda^{k+1}}.$$
Now if we multiply this formula by $\dfrac{e^{\lambda t}}{2\pi i} $ and path-integrate along a circle $C_r$ with radius $r$ such that $r>|A|$, using the formula $ \frac{1}{2\pi i}\int_{C_r}\frac{e^{\lambda t}}{\lambda^{k+1}}d\lambda =\frac{t^k}{k!}$ we find
$$\frac{1}{2\pi i}\int_{C_r}e^{\lambda t}(\lambda I-A)^{-1}d\lambda=\sum_{k=0}^{\infty}\left( \frac{1}{2\pi i}\int_{C_r}\frac{e^{\lambda t}}{\lambda^{k+1}}d\lambda \right)A^k=\sum_{k=0}^{\infty}\frac{t^k}{k!}A^k=e^{tA}.$$
I am just not sure how I can justify the fact that I interchanged the series and integral. And I don't know how I can use Cauchy's integral formula to deform the circle $C_r$ to the line $L_a=\{\lambda\in \mathbb{C}, \ \ Re \ \lambda=a \}$ where $a\geq r$ so I can have
$$e^{tA}=\frac{1}{2\pi i}\int_{L_a}e^{\lambda t}(\lambda I-A)^{-1}d\lambda.$$
 A: If $t\gt0$, then we want $a\ge2|A|$ and 
$$
\gamma_r=a+ir[-1,1]\cup[a,-r]+ir\cup-r+ir[1,-1]\cup[-r,a]-ir
$$
If $t\lt0$, then we want $a\le-2|A|$ and
$$
\gamma_r=a+ir[1,-1]\cup[a,r]-ir\cup r+ir[-1,1]\cup[r,a]+ir
$$
In each case, $\gamma_r$ is counterclockwise.
When $t\gt0$, the finite part of $\gamma_r$ is $[a-i\infty,a+i\infty]$.
When $t\lt0$, the finite part of $\gamma_r$ is $[a+i\infty,a-i\infty]$.
Since $\lambda I-A$ is invertible along $\gamma_r$, where $r\gt|A|$, we have
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}\frac1{2\pi i}\int_{\gamma_r}e^{\lambda t}(\lambda I-A)^{-1}\mathrm{d}\lambda
&=\frac1{2\pi i}\int_{\gamma_r}\lambda e^{\lambda t}(\lambda I-A)^{-1}\mathrm{d}\lambda\\
&=\frac1{2\pi i}\int_{\gamma_r}(\lambda I-A+A)e^{\lambda t}(\lambda I-A)^{-1}\mathrm{d}\lambda\\
&=\frac1{2\pi i}\int_{\gamma_r}Ae^{\lambda t}(\lambda I-A)^{-1}\mathrm{d}\lambda
+\frac1{2\pi i}\int_{\gamma_r}e^{\lambda t}\mathrm{d}\lambda\\
&=A\frac1{2\pi i}\int_{\gamma_r}e^{\lambda t}(\lambda I-A)^{-1}\mathrm{d}\lambda
\end{align}
$$
Therefore, for some $C$, we get
$$
\frac1{2\pi i}\int_{\gamma_r}e^{\lambda t}(\lambda I-A)^{-1}\mathrm{d}\lambda=Ce^{tA}
$$
where
$$
\begin{align}
C
&=\frac1{2\pi i}\int_{\gamma_r}(\lambda I-A)^{-1}\mathrm{d}\lambda\\
&=\frac1{2\pi i}\int_{\gamma_r}\frac1\lambda\left(I-\frac A\lambda\right)^{-1}\mathrm{d}\lambda\\
&=\frac1{2\pi i}\int_{\gamma_r}\left(\frac I\lambda+\frac A{\lambda^2}+\dots\right)\mathrm{d}\lambda\\
&=I
\end{align}
$$
Thus,
$$
\frac1{2\pi i}\int_{\gamma_r}e^{\lambda t}(\lambda I-A)^{-1}\mathrm{d}\lambda=e^{tA}
$$
A: You can reduce integral identities to the scalar integrals by applying the operators to vectors and then applying a bounded linear functional $x^{\star}$ to the corresponding vector expressions. Because the bounded linear functionals separate points, you can later remove them from your expressions to obtain a vector identity; finally the vectors are removed to obtain an operator identity. About the only thing you need to know is
$$
        x^{\star}\left(\int_{C} F(\lambda)\,d\lambda\,x\right)=\int_{C}x^{\star}(F(\lambda)x)\,d\lambda.
$$
The estimate that you need to convert your integral to a line integral is
$$
            \|(\lambda I -A)^{-1}\| = \| \sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}A^{n}\|
       \le \sum_{n=0}^{\infty}\frac{\|A\|^{n}}{|\lambda|^{n+1}}=\frac{1}{|\lambda|-\|A\|},
$$
which is valid for $|\lambda| \ge \|A\|$.
